current voltage relationship of inductor (are the two orthogonal)

Discussion in 'General Electronics Chat' started by u04f061, Oct 16, 2006.

1. u04f061 Thread Starter Member

Jan 3, 2006
14
0
the following signals are known to be orthogonal

i(t)=k(exp(jωt))
v(t)=kjω(exp(jωt))

we can take i(t) is the current through inductor and v(t) as the voltage across inductor. in the book of b.plathi (modern digital and analog comn systems) it is written that two complex signals are orthogonal if integral of product of one signal with the conjugate of other is zero. that is

∫i(t)*v(t)dt=0 where * is the symbol showing that v(t) is conjugated.

but the calculations i did show that this untegral is non-zero. plz have a look at my caclculations

∫i(t)*v(t)dt

=k²(-j)ω∫exp(jωt)exp(-jωt)
=-k²jω∫dt
≠0

henc this relation shows that these signals are not orthogonal.

2. Papabravo Expert

Feb 24, 2006
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I'm not sure why you think that the two expressions should be orthogonal. It is true that one expression is the derivative of the other, but I'm not aware of any theorem that says that a function and it's derivative must be orthogonal.

In a physical sense I see no reason why voltage and current as functions of time need to be orthogonal. Especially since the ratio at a given frequency is the impedance or admittance.

3. Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
I think you missed his point. The voltage and current in an ideal inductor are orthogonal. That is a fact. I think what our u04f061 is questioning is the definition in the book he quoted, or in his evaluation of that definition.
The simple definition of orthogonality is that if the dot product of two functions is zero, the functions are orthogonal. This is clearly true of the voltage and current in an inductor.
I don't know about the definition in b. plathi's book. I couldn't find anything wrong with u04f061's math, but I'm no great mathematician.

4. Papabravo Expert

Feb 24, 2006
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If the current and voltage are orthogonal in an ideal inductor then the two original expressions cannot be for the voltage and the current. The problem may be in the sloppy formulation of the integral. He starts with an expression that does not show the variable that the integration is to be performed under. If you have a function of a complex variable then don't you need to make the integral a path integral?

5. Dave Retired Moderator

Nov 17, 2003
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I am not convinced that the conjugate of v(t)=kjω(exp(jωt)) is v(t)= -kjω(exp(-jωt)). Where has the minus sign in the exponential power come from?

Dave

6. u04f061 Thread Starter Member

Jan 3, 2006
14
0
oh sorry bcoz i forgot to write integration limits. u can clearly visualize that exponential is an analytic function so it does not require any path. it only requires integration limits. in all the paths the result of integral would be same for a particular limit of t say [t1,t2] where t2>t1
plz have a review of cauchy integral theorem.

second thing u were saying that it does not include any variable over ,plz have a look at dt

7. u04f061 Thread Starter Member

Jan 3, 2006
14
0
plz have a review of ac analysis of your basic circuit course.in that case we plot real and complex variables along two orthogonal vectors (cartesian coordinate system)

8. Dave Retired Moderator

Nov 17, 2003
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172
Having played around with Matlab, I can confirm that the conjugate of v(t)=kjω(exp(jωt)) is v(t)*= -kjω(exp(-jωt))

Assuming k = ω = t = 1:

v(t) = -0.84147 + 0.5403j

and

v(t)* = -0.84147 - 0.5403j

Verifyable by the conj function.

Dave

9. u04f061 Thread Starter Member

Jan 3, 2006
14
0
yah it is clear fact that to take complex conjugate of a function we need only to invert j (imaginary ) components

10. Dave Retired Moderator

Nov 17, 2003
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I know. My confusion arose out of the fact that there was an imaginary component within the exponential power. As it happens this is irrelevant for the purposes of calculating a conjugate.

Dave

11. n9352527 AAC Fanatic!

Oct 14, 2005
1,198
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I agree with Papabravo in this case, it seems that the two original functions for current and voltage are not orthogonal.

It is true that the voltage across an ideal inductor is the derivative of the current, as given by V=LdI/dt. However, I don't think just by obtaining the voltage function from the derivative of the current function make the two resulting functions orthogonal.

I couldn't find anything wrong with u04f061's calculation and integration. These can be verified by using the relationship exp(jwt) = cos(wt) + j sin(wt) and by going through the same treatments as his would also give the same result.

12. Papabravo Expert

Feb 24, 2006
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The j in the exponent certainly has an effect if you use the expansion
exp(jwt) = cos(wt) + jsin(wt)
The conjugate is clearly
conj(exp(jwt)) = cos(wt) -jsin(wt) = cos(-wt) + jsin(-wt) = exp(-jwt)
using the properties of odd and even functions namely
cos(-x) = cos(x) -> even function
sin(-x) = -sin(x) -> odd function

Feb 24, 2006
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